Ok I've got to evaluate the following limit:
But I've got no idea how the absolute values can be rewritten in this limit?
Note that $\displaystyle \left|x-1\right|=\left\{\begin{array}{rl} x-1, & \text{ if }x\geq 1\\ {\color{red}-x+1}, & \text{ if }x<1\end{array}\right.$ and $\displaystyle \left|x+1\right|=\left\{\begin{array}{rl} {\color{red}x+1}, & \text{ if }x\geq -1\\ -x-1, & \text{ if }x<-1\end{array}\right.$
Thus, $\displaystyle \lim_{x\to 0}\frac{x}{\left|x-1\right|-\left|x+1\right|}=\lim_{x\to 0}\frac{x}{(-x+1)-(x+1)}=\lim_{x\to0}-\frac{x}{2x}$.
I'm sure you can take it from here.
Thanks very much for illustrating Chris, but I have no idea how you chose x-1 in the 2nd set of brackets in $\displaystyle \lim_{x\to 0}\frac{x}{(-x-1)-(x-1)}$???
And why the red ones in particular? I'm not sure which set of arbitrary values this limit should be taking? This would be easier with one sided limits.
By the way, how do you know which of the greater than or less than signs you should put the equals sign under?
i.e. how can you tell which sign needs to have the 'or equal' with it because attaching the equals under greater than or less than are equally possible?